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Calculate the empirical formula the compound with the following percent composition: 27.59%C 1.15%H 16.09%N 55.17%O

Respuesta :

danielesartori

Answer: C2HNO3

Explanation: C = 27.59/12.011 = 2.297

H = 1.15/1.008 = 1.1409

N = 16.09/14.007 = 1.1487

O = 55.17/15.999 = 3.4483

Divide by smallest result:

C = 2

H=1

N=1

O = 3

Empirical formula = C2HNO3

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